Talk:Sir Finley Mrrgglton

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Can your current basic Hero Power appear in the Discover selection?[edit]

I think this card's Discover effect cannot select your current Hero Power as one of the three randomly selected basic Hero Powers: is this confirmed? In both cases, I suggest to mention this in the "Notes" section. --Elekim (talk) 08:52, 10 January 2016 (UTC)

AFAIK it cannot, but I'd have to look for a source to back me up. --Patashu (talk) 00:28, 11 January 2016 (UTC)
You added a note confirming it cannot appear to the Notes section, but didn't add a reference. I take it you found a source? -- Taohinton (talk) 16:27, 22 January 2016 (UTC)
Late, but I have a source now. --Patashu (talk) 03:41, 25 January 2016 (UTC)
The extra info is nice - answers a question I was wanting to ask ;) -- Taohinton (talk) 04:05, 25 January 2016 (UTC)

Classes' non-standard Hero Powers[edit]

So, Lesser Heal will not be offered to you by Sir Finley Mrrgglton if you have Mind Shatter (even if you are not a Priest) since both are considered to be "Priest Hero Powers". Similarly, is Lightning Jolt considered a Shaman Hero Power, and is The Tidal Hand a Paladin Hero Power? Elekim (talk) 06:49, 19 June 2016 (UTC)

Yes and yes.
[09:03] <boombot> The Tidal Hand [OG_006b OG][Paladin Hero_power][2 mana]: Hero Power Summon a 1/1 Silver Hand Murloc.
[09:04] <boombot> Lightning Jolt [AT_050t TGT][Shaman Hero_power][2 mana]: Hero Power Deal 2 damage. --Patashu (talk) 23:04, 20 June 2016 (UTC)

Kara trailer[edit]

I just had to note that the trivia about the "hot mana font" User:Taohinton recently added to this page might be my favorite wiki edit of all time :) Though to be fair, any mention of random cows is probably gonna get a laugh out of me... - jerodast (talk) 15:48, 31 July 2016 (UTC)

Probability of getting specific hero power with sir finley[edit]

I'm pretty sure the probability calculation is wrong. It is based on the idea that each of the three slots have a one in eight chance of being the one you want but I think that's faulty.

Say your hero is Druid and you want Life Tap. Let's go through the slots one by one:

  • The first slot does indeed have a 1/8 chance of being life tap. Let's say it isn't, it's Lesser Heal.
  • The second slot now has a 1/7 chance of being Life Tap. Seeing as it can't be Shapeshift or Lesser Heal (each hero power only appears once, right?). Alas, it's Armor Up so we move on to the third.
  • The third slot then obviously has a 1/6 chance of being Life Tap.

So the total probability of getting a particular hero power would be the sum of these probabilities:

1/8 + 1/7 + 1/6 = 44%

Right? — Preceding unsigned comment added by (talkcontribs) 15:10, 31 August 2016‎

Valid considerations but it turns out it still works out to 3/8. This is an extremely easy mistake to make, and a difficult one to explain. Let me offer THREE perspectives on this, because I have no life:
The correct "dependent events" probabilities
  1. You have a 1/8 chance on Life Tap the first time.
  2. You have a 1/7 chance on Life Tap the second time, but only if you didn't get it the first time, which has a 7/8 probability. When you have one probability conditional on another, you multiply. So the chance of Life Tap the second time is 1/7 * 7/8 = 1/8.
  3. So the chance of getting Life Tap the first two times is 1/8 + 1/8 = 2/8.
  4. You have a 1/6 chance on Life Tap the third time, but only if you didn't get it the first two times, which has a 6/8 probability. You multiply, and get 1/6 * 6/8 = 1/8.
So the chance of getting Life Tap any of the first three times is 1/8 + 1/8 + 1/8 = 3/8.
Obviously the part most people forget is that you can only count the "higher" chance the second time IF the second time actually had a chance of improving on the first. Since the improved probability the second time is CONDITIONAL on the first pick going AGAINST you (otherwise the second pick would be completely irrelevant!), you must factor that in. You can't just "double count" the full 1/7 probability for both the situation where you didn't draw it the first time AND the situation where you did.
It is easy to see this would be wrong with a lower pool size. Suppose we have only 5 total powers, so in terms of replacements, it's from a pool of 4. Then by the proposed logic, the chances would be 1/4 for the first pick, 1/3, and 1/2 for the others. Adding, we get 25% + 33% + 50% = 108% chance of choosing the desired one! Yet clearly, there should be a chance you failed, and it shouldn't be over 100%. This is because we did not observe that 33% only applied if we missed the 25%, and so forth for any future picks.
The "single event" perspective
In general, it turns out that the perspective of "one event after another" or "a single event with multiple parts" makes no difference to the final probability. Yes, the computer is very likely determining one power, then another, then a last one, each time reducing the pool. Much like you'd pick from a deck of cards. However, what if instead of cards you were asked to reach into a bag of 8 marbles and grab 3 at the same time? It's not like your index finger chooses one, then your thumb, then your middle finger. You simply make a grab, and now you have 3 marbles. (If you get too many/few you try again.) 3 in your hand, 5 in the bag, so pretty clearly the chance of any one marble being in there is 3/8. Is the concept of picking 3 marbles out of a bag fundamentally different from picking 3 cards out of a deck? Nope.
The probability space
Ultimately a lot of probability can be broken down to simply listing every individual outcome, and then counting how many of those apply to your conditions. This set of outcomes is called the probability space. In our simplified "size 4 power pool" example, here are ALL options, with numbers indicating which of 4 powers were chosen in what order:
  • 1, 2, 3 -- 1, 3, 2 -- 2, 1, 3 -- 2, 3, 1 -- 3, 1, 2 -- 3, 2, 1 (6 ways to choose 1, 2, 3)
  • 1, 2, 4 -- 1, 4, 2 -- 2, 1, 4 -- 2, 4, 1 -- 4, 1, 2 -- 4, 2, 1 (6 ways to choose 1, 2, 4)
  • 1, 3, 4 -- 1, 4, 3 -- 3, 1, 4 -- 3, 4, 1 -- 4, 1, 3 -- 4, 3, 1 (6 ways to choose 1, 3, 4)
  • 2, 3, 4 -- 2, 4, 3 -- 3, 2, 4 -- 3, 4, 2 -- 4, 2, 3 -- 4, 3, 2 (6 ways to choose 2, 3, 4)
Probability calculations relies on the axiom that any one of these individual sequences is equally likely. With this arrangement it's clear that for any desired number, only 6/24 = 1/4 of the options will fail to include that number, and 3/4 of them will include it. If you add a 5th, you'd see 2/5 and 3/5, and for 8 options, you see 5/8 failing to include, and 3/8 including the number. You could view each selection of a number as narrowing down the "space" you're looking in, OR you can just count how many of the final results match your conditions, since they're all equally likely.
This is way more than anyone was asking for I know, but maybe we can link back to it if someone else has questions. Or maybe we should make a little primer on probability somewhere :) It can be complex but is obviously pretty relevant to card games and RNG! - jerodast (talk) 09:37, 1 September 2016 (UTC)
I was wanting to reply with more or less the same explanation but couldn't find the words or energy at the time - now I'm glad I left it to you! xD In terms of the final paragraph, this is certainly something worthy of a page on the wiki, and you seem like the right person to write it :P I would suggest focusing on showing readers how to calculate probabilities for a given card or effect, so that they can then apply that to any game situation they come across. We already have RNG, but that's more of a discussion page (from the early days when everyone was still complaining Hearthstone was purely luck-based). Maybe Probability, or else we can use that as a redirect for convenient linking? -- Taohinton (talk) 14:11, 1 September 2016 (UTC)